Given real positive numbers a,b,c and d, for which the equalities a2+ab+b2=3c2 and a3+a2b+ab2+b3=4d3 are fulfilled. Prove that a+b+d≤3c.
Problem
Source: 2016 Latvia BW TST P5
Tags: algebra, inequalities
17.12.2022 14:12
WLOG a+b=1. Let ab=m. Then 12−m=3c2 and 13−2×1⋅m=4d3. So {1−m=3c2,1−2m=4d3. Meaning that 6c2=4d3+1. Notice thatab=m≤(a+b2)2=14,so 4d3≥1−14×2, so d≥12. Our inequality becomes 1+d≤3c. Squaring and substituting implies that proving 12d3−2d2−4d+1≥0should suffice. Factorize this into (2d−1)(6d2+2d−1) and we see that it is true.
17.12.2022 15:03
youthdoo wrote: WLOG a+b=1. Let ab=m. Then 12−m=3c2 and 13−2×1⋅m=4d3. So {1−m=3c2,1−2m=4d3. Meaning that 6c2=4d3+1. Notice thatab=m≤(a+b2)2=14,so 4d3≥1−14×2, so d≥12. Our inequality becomes 1+d≤3c. Squaring and substituting implies that proving 12d3−2d2−4d+1≥0should suffice. Factorize this into (2d−1)(6d2+2d−1) and we see that it is true. Why could you assume a+b=1? Is there a name for this technique?
17.12.2022 15:08
Because the variables are to the same power in each equation or inequality.
17.12.2022 19:00
parmenides51 wrote: Given real positive numbers a,b,c and d, for which the equalities a2+ab+b2=3c2 and a3+a2b+ab2+b3=4d3 are fulfilled. Prove that a+b+d≤3c. Reducing c and d, the inequality to prove can be written as a+b+3√(a+b)(a2+b2)4≤√3(a2+ab+b2) Using Holder's inequality and 34(a+b)2≤a2+ab+b2: 2⋅a+b2+3√(a+b)(a2+b2)4≤3√9[(a+b)38+(a+b)38+(a+b)(a2+b2)4]=3√94(a+b)[(a+b)2+a2+b2]=3√92(a+b)(a2+ab+b2)≤√3(a2+ab+b2)